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Q. Suppose S is a bounded and closed nonempty subset of real numbers. Prove $\sup S$ is in $S$.

Since $S$ is bounded and by the least upper bound property of $\mathbb R$, there exists $\sup S \in \mathbb R$ which we do not know if it is in $S$ yet.

By the definition of supremum, for $\epsilon \gt 0$,

$\exists s \in S$ such that, $\sup S - \epsilon \lt s \lt \sup S + \epsilon$

So $s$ is in the interval, $(\sup S - \epsilon, \sup S + \epsilon)$

Then $S \cap (\sup S - \epsilon, \sup S + \epsilon) \neq \emptyset$

Then $\Bigl( (\sup S - \epsilon, \sup S + \epsilon) \setminus \{\sup S\} \Bigr) \cap S \neq \emptyset$

We also know that a closed set contains all of its accumulation points and since $\sup S$ is an accumulation point of $S$, we proved $\sup S \in S$


Does this proof look okay? If not, can you give me hints on where I went wrong?

Thank you.

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It is almost fine, but not entirely. You jumped from$$S\cap(\sup S-\varepsilon,\sup S+\varepsilon)\neq\emptyset$$to$$S\cap\bigl((\sup S-\varepsilon,\sup S+\varepsilon)\setminus\{\sup S\}\bigr)\neq\emptyset\tag1$$without any justification. If it turns out that $s=\sup S$, $(1)$ may well be false.

This is easy to solve, though. Just begin by saying that this is a proof by contradiction. That is, you assume that $\sup S\notin S$. Then $(1)$ would be true and you reach a contradiction when you arive that $\sup S\in S$ after all.

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  • $\begingroup$ I'm quite confused. Can you elaborate more on if I assume that sup $S \notin S$, then (1) would be true? thank you $\endgroup$ – TUC Oct 28 '18 at 9:52
  • $\begingroup$ You know that $S\cap(\sup S-\varepsilon,\sup S+\varepsilon)\neq\emptyset$ because you proved that there is some $s\in S\cap(\sup S-\varepsilon,\sup S+\varepsilon)$. But if that $s$ is precisely $\sup S$, then how do you know that $(1)$ holds? The only difference between $S\cap(\sup S-\varepsilon,\sup S+\varepsilon)$ and $S\cap\bigl((\sup S-\varepsilon,\sup S+\varepsilon)\setminus\{\sup S\}\bigr)$ is that $\sup S$ belongs to the first set, but not to the second one. $\endgroup$ – José Carlos Santos Oct 28 '18 at 10:00
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At the beginning, you better mention that your S is (non-empty) and bounded so there is a real supremum.

Otherwise, your proof is correct.

You have explained every step very clearly. You may as well make a proof based on sequences which is basically the same idea.

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  • $\begingroup$ Every closed and bounded set $S$ in $\Bbb{R}$ is compact. Since $\Bbb{R}$ is a metric space, $S$ is also sequentially compact. In particular, a sequence $\{ x_n \}$ of points in $S$ such that $x_{n+1} \ge x_n$ has a subsequence that converges in $S$ and it must be the supremum. Is this a right approach to the sequence proof you mentioned? $\endgroup$ – Niki Di Giano Oct 28 '18 at 9:32
  • $\begingroup$ That is correct. Thanks $\endgroup$ – Mohammad Riazi-Kermani Oct 28 '18 at 13:28

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